1) The final velocity of the object is -78 m/s (downward)
2) The final height of the object is -290 m
Explanation:
1)
When the object is thrown up, it follows a free fall motion, subjected only to the force of gravity (that pulls the object downward). Therefore, the object has a constant acceleration
[tex]g=9.8 m/s^2[/tex]
towards the ground (acceleration of gravity).
Since this is a motion with constant acceleration, we can use the following suvat equation:
[tex]v=u+at[/tex]
where
v is the final velocity
u is the initial velocity
a = g is the acceleration
t is the time of flight
In this problem we have:
u = +20 m/s (taking upward as positive direction)
t = 10 s is the time of flight
[tex]g=-9.8 m/s^2[/tex] (downward, so negative)
Solving for v, we find the final velocity:
[tex]v=20+(-9.8)(10)=-78 m/s[/tex]
2)
For this part of the problem, we can use another suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
u is the initial velocity
a is the acceleration
t is the time
s is the displacement of the object
In this problem,
u = +20 m/s
t = 10 s
[tex]g=-9.8 m/s^2[/tex]
Substituting, we find the final position of the object with respect to the initial position:
[tex]s=(20)(10)+\frac{1}{2}(-9.8)(10)^2=-290 m[/tex]
Learn more about uniform accelerated motion:
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